Ultimate effort on a beam in bending

[Vincent73]

Hi,

I would like to calculate the ultimate load on a cylindrical beam (length 70mm, diameter 12mm) fixed at an extremity, and a ponctual load applied at the other extremity.
My research allowed me to calculate the effort at yield limit (elastic modulus), and the effort when all the section has plastified (plastic section modulus).

However, those criteria do not correspond to the ultimate strength of the beam. After making some tests, it appears that the effort at the ultimate strength is higher than the maximum effort calculated via plastic section modulus.

Do you know how to calculate the ultimate effort on a beam (I mean the real maximum effort that the beam is able to support).

Thanks a lot!

 

[rb1957]

"ponctual" ? do you mean point (load) ? or distributed over the end face ? (effectively the same in analysis).

the beam you describe is a cantilever, the analysis you describe sounds reasonable. How far out is the test result ? Can you have material coupons form the test specimen to verify properties ?

another day in paradise, or is paradise one day closer ?

 

[Vincent73]

Thanks for your answer rb1957.

Yes it is point load.
Concerning the material, I have material certificate so it should be correct.

For me, the critiria with the use of plastic section modulus corresponds to a section that plastifies entirely.
But my interpretation is that it does not correspond to the ultimate strengh (top of the stres/strain curve before the failure) of the beam.

It is a shame because i can calculate it correctly if the test is a tensile test (using Rm : Ultimate strength allowable data), but I don't know how to do it with a bending test.

What do you think about that ?

thanks

 

[rb1957]

the section should be completely plastic at failure … assuming that it is not failing early by compression crippling.

by "a section that plastifies entirely" do you mean setting the stress to fty (+ve and -ve). This is conservative, ie short of actual.

you can do better with "Cozzone" analysis, which allows some stress to exceed fty. You can read about more detailed analysis for Modulus of Rupture in Bruhn "analysis of Flight Vehicle Structures", sect C3.

another day in paradise, or is paradise one day closer ?

 

[dhengr]

Vincent73:
You have to define ultimate strength very carefully, and also understand what that really means. The meaning might be slightly different depending upon the exact problem, or the stress picture, stress fields involved. And, you have to understand these basic differences if you want to be a good Structural Engineer. In the tension test, which we all see/learn about early in our careers/studies, we can clearly see elastic behavior with applied tensile loading, then we see inelastic behavior or plastification and/or work hardening, all on a pretty uniform stress field, and finally with increased load application we see ultimate load/strength, or failure and specimen separation. And, this is a pretty clear-cut picture, stress wise and limit wise.

With your cantilevered beam problem, you have quite a different problem, much different stress conditions, a short enough beam vs. dia. (depth) so you likely have reaction end and loaded end localized stress conditions/anomalies too. As you apply an increasing end load, you can see elastic bending and you know about where it is at its max., and you can plot deflections, and unloading the beam you can see that it is still acting elastically. Then, in the case of most steels, the outer fibers start to go beyond the yield strength of the material, and you can see that the beam does not return to zero when unloaded. It is starting to form a plastic hinge, and increased loading increases the hinging size/action and the deflections. At some loading the hinge area of the beam kinda becomes like a wet noodle (a true hinge), and deflection grows swiftly, while the beam won’t carry any more load. At this point the beam has failed, while not separating and falling on someone’s head. There is no ultimate strength or load, in the same sense as in the tension test, but the beam has reached the limit of its usefulness, and the saving grace is that the beam did not fail/separate and hurt anyone. For this reason you will see that conditions which lead to sudden disastrous failures (like tension tested or tension rods or eye bars, and the like) are treated with a higher factor of safety and other limiting restrictions to prevent the failure without some obvious warning. We usually set the stress/strain limit some percentage of the way up the strain hardening curve to intentionally stay away from the ultimate/breaking limit.

 

[Vincent73]

Thanks rb1957, dhengr.

Dhengr you are true, I agree with all you said. You are right also to remind the ration L/10 and small displacement theory etc..
To be more precise then: I'm trying to determine the load when the beam won't carry anay more load (it could be without any visible crack/failure).
So the value i'm looking for is not when the beam is separating in two parts, but when it can not carry any more load.

My application is for building industry.
Usually for this application, I use Eurocode Elastic criteria for Service Limite State (with a safety coefficient), and plastic section modulus criteria (with a safety coefficient) and Ultimate Limit State 1.5 if it's the wind for example. Results show that the capacity to carry loads of the beam tested can go something like 30% higher than the calculation with plastic section modulus (without safety coefficient). The number of tests done is 10, and the repeatability quite good. Also verifying buckling & lateral buckling accoding to standards.

For the safety part I have no doubt the design is OK, but I need to know... How to calculate the real limit this beam can reach?! It's often a question comming from commercials or people who don't know much about theorical aspects, and the question is so simple that they wait a fast and simple answer. And I think they are right because Elastic/Plastic is not words everyone can understand.

Here is the difference between elastic section modulus and plastic section modulus well illustrated Link.

For a tensile test, with material data (Stress at ultimate strength) you can calculate the load when the beam can not carry any more load by doing:
Sigma_Ultimate = Load/Area (Area is reduced and can be calculated with Poisson's ratio).

By reading your comment, I have now a stong conviction that it can not be calculated easily for a beam in bending. Because outer fibers may have failed but inner fibers could always be in elastic domain, but this will not necessarily mean that the beam is not able to carry more load.
Also the second moment of area involved in the calculation has always been for me difficult to undertand: I mean let me alone in a room on desk with a paper and a pen, and suppose i forgot the formulation of the moment of area: it will be quite impossible for me to build this formulation alone. So if i'm not able to build the theory of elastic bending, I will not for failure bending theory...

Thanks rb1957, I'm gonna read Cozzone analysis, it looks great. Also I will check with Bruhn if I can find a second hand book.
Sorry I don't know the meaning of fty (I suppose it is ft.. yield ?)

 

[rb1957]

I get you're after some "truth", as you say it's a simple question that people ask.

An answer I'd give would be our plastic analysis, saying this is the load we can be sure, reasonably sure, the beam can carry before failure. Codes are written understanding this level of analysis/certainty/definition of the structure. Sure, we think that often the real failure load is some 30% higher, but that is "just" some safety room hidden in the code. If we spend more time and calculated more carefully and based our analysis on the "true" strength of the beam, well …
1) we'd lose a level of safety of our code
2) we're closer but still not definitely at The failure load of the beam, since material properties change/vary
3) if we do So much work as to predict the true failure (to within a few %age points) of the actual beam we were building, I think we'd run out of customers (there's so much work involved in answering simple questions like this).

another day in paradise, or is paradise one day closer ?

 

[dhengr]

Vincent73:
You will likely find that those kinds of questions, from the general public, be it in a courtroom or at a town meeting, etc., or even from your own management at times, are often better answered at their own level of understand of the subject matter. That tends to be late grade school, early high school science, physics and chemistry level, for much of the general public, and that includes a lot of Uni. grads too. I mean no particular disrespect, but you don’t have to get to involved or complicated before people’s eyes start to glaze over when explaining things scientific and engineering wise. They usually aren’t impressed with fancy/elaborate calcs. which supposedly pull out the last 5-10% of ultimate strength, whatever that really means, they want an answer that they can visualize and kinda understand given their background and experiences. Simple real life examples, simple experiments to show concepts and the like often satisfy far better than your trying to claim you might get another few pounds out of the beam, you probably don’t even know the exact load conditions that well. Sometimes, it might be better to answer with another question…, ‘do you really want it to break and hurt or kill you or someone else?’ Thus, we take our calcs. and designs to a serviceability limit, a useful life limit, the bldg. code limits, in our best engineering judgement, and don’t tempt fate by trying to utilize the last few %, which, by the way, becomes less and less certain for the many reasons given above. Explain some of the uncertainties which we must deal with and design around, as regards materials variables, exact loading uncertainties, manufacturing and construction variations, and why we apply a factor of safety in our work, for the public’s safety. Use your own example of the tensile test, and a brief description of the test, and the sudden failure, which we can not tolerate in real life, and the simple beam which starts to form a hinge and deflects too much to be useful, but warns us of imminent failure problems, but doesn’t hurt anyone.

 

[GregLocock]

I can think of reasons why you might get a lower answer than predicted by plastic theory, higher is a much more interesting case. For my final year project the workshop built steel trusses to fail in various ways, and then tested them, and we tried to explain the results (force deflection curves up to and after failure - wrecked a few clocks). It turned out we had been taught sufficient second order theory to figure out the effect of softening as joints approached yield, but 35 years later the details are a mist.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?